Question

A cyclist riding the bicycle at a speed of $14\sqrt{3}m{s}^{-1}$ takes a turn around a circular road of radius $20\sqrt{3}$ m without skidding. Given $g=9.8m{s}^{-2}$, what is his inclination to the vertical m

• A. ${30}^{\circ }$
• B. ${90}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is D

${60}^{\circ }$

$\theta =ta{n}^{-1}\left(\frac{v^2}{rg}\right)=ta{n}^{-1}\left[\frac{(14\sqrt{3}{)}^2}{20\sqrt{3}\times 9.8}\right]=ta{n}^{-1}\left[\sqrt{3}\right]={60}^{\circ }$